Put the equation in vertex form: y^2=12x-2y+250 using the formula y=a(x-h)^2+k
Thanks in advance :]
3 answers
Oops, its y^2 - 12x - 2y + 25 = 0
We would not state your parabola in the form
y = a(x-h)^2 + k, because your parabola has a horizontal axis of symmetry (lays sideways)
anyway ....
12x = y^2 + 2y - 250
12x = y^2 + 2y + 1 - 1 - 250
12x = (y+1)^2 - 251
x = (1/12)(y+1)^2 - 251/12
notice the similarity between the parabola equations, the x and y variables have been interchanged. So to state the vertex we also have to
interchange the variables.
the vertex is (-251/12, -1)
check my arithmetic
y = a(x-h)^2 + k, because your parabola has a horizontal axis of symmetry (lays sideways)
anyway ....
12x = y^2 + 2y - 250
12x = y^2 + 2y + 1 - 1 - 250
12x = (y+1)^2 - 251
x = (1/12)(y+1)^2 - 251/12
notice the similarity between the parabola equations, the x and y variables have been interchanged. So to state the vertex we also have to
interchange the variables.
the vertex is (-251/12, -1)
check my arithmetic
12x = y^2 - 2y + 25
12x = y^2 - 2y + 1 - 1 + 25
12x = (y-1)^2 + 24
x = (1/12)(y-1)^2 + 2
vertex is (2, 1)
12x = y^2 - 2y + 1 - 1 + 25
12x = (y-1)^2 + 24
x = (1/12)(y-1)^2 + 2
vertex is (2, 1)